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- Feb 14, 2012

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Find the value of $\displaystyle \frac{1^4+2007^4+2008^4}{1^2+2007^2+2008^2}$.

My Attempt:

For the numerator, I expand them using the Binomial Theorem and get:

$\displaystyle 1^4+2007^4+2008^4$

$\displaystyle=1+(2000+7)^4+(2000+8)^4$

$\displaystyle

=1+2000^4+4(2000)^3(7)^1+6(2000)^2(7)^2+4(2000)^1(7)^3+7^4$

$\displaystyle

+2000^4+4(2000)^3(8)^1+6(2000)^2(8)^2+4(2000)^1(8)^3+8^4$

$\displaystyle

=2(2000^4)+60(2000)^3+678(2000)^2+3420(2000)+6498$

I do the same for the denominator and get:

$\displaystyle 1^2+2007^2+2008^2$

$\displaystyle=1+(2000+7)^2+(2000+8)^2$

$\displaystyle=1+(2000)^2+14(2000)+49+(2000)^2+16(2000)+64$

$\displaystyle=2(2000^2)+30(2000)+114$

I noticed that if I square the denominator, I get twice the value as the numerator, let's see...

$\displaystyle (1^2+2007^2+2008^2)^2$

$\displaystyle

=4(2000^4)+120(2000)^3+1356(2000)^2+6840(2000)+12996$

$\displaystyle

=2(2(2000^4)+60(2000)^3+678(2000)^2+3420(2000)+6498)$

So, we can say that

$\displaystyle \frac{1^4+2007^4+2008^4}{(1^2+2007^2+2008^2)^2}= \frac{1}{2}$

Therefore,

$\displaystyle \frac{1^4+2007^4+2008^4}{1^2+2007^2+2008^2}=\frac{1}{2}(1^2+2007^2+2008^2)=\frac{1}{2} (2(2000^2)+30(2000)+114)=4,030,057$

My question is, are there any other approaches to solve this problem?

Thanks in advance.